Two radioactive materials A and B have decay | vedclass

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(C) Given: Decay constants $\lambda_{A} = 5\lambda$ and $\lambda_{B} = \lambda$.At $t=0$,the initial number of nuclei are equal,i.e.,$(N_{0})_{A} = (N

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$\frac{1}{2\lambda}$

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(C) Given: Decay constants $\lambda_{A} = 5\lambda$ and $\lambda_{B} = \lambda$.At $t=0$,the initial number of nuclei are equal,i.e.,$(N_{0})_{A} = (N_{0})_{B}$.We are given the ratio $\frac{N_{A}}{N_{B}} = (\frac{1}{e})^{2} = e^{-2}$.According to the law of radioactive decay,$N = N_{0}e^{-\lambda t}$.For material $A$,$N_{A} = (N_{0})_{A} e^{-5\lambda t}$.For material $B$,$N_{B} = (N_{0})_{B} e^{-\lambda t}$.Dividing the two equations:$\frac{N_{A}}{N_{B}} = \frac{(N_{0})_{A} e^{-5\lambda t}}{(N_{0})_{B} e^{-\lambda t}} = e^{-(5\lambda - \lambda)t} = e^{-4\lambda t}$.Equating the ratios:$e^{-4\lambda t} = e^{-2}$.Comparing the exponents:$4\lambda t = 2$.Solving for $t$:$t = \frac{2}{4\lambda} = \frac{1}{2\lambda}$.

Two radioactive materials $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t=0$,they have the same number of nuclei. The ratio of the number of nuclei of $A$ to that of $B$ will be $(1/e)^2$ after a time interval of:

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